Group Theory: Cyclic groups and Subgroup of Dihedrals, is my proof okay? Suppose we're given a Dihedral group $D_{30}$.
a) Find a cyclic subgroup H of order 10 in $D_{30}$. List all generators of H.
b) Let k and n be an integer such that k >= 3 and k divides n. Prove that $D_n$ contains exactly one cyclic subgroup of order k.  
My attempt:
a) in $D_{30}$ we know that |r| = 30. So we can find a cyclic subgroup H generated by r such that it is of order 10. take < $r^{30/10}$ > = < $r^3$ >. Then < $r^3$ > contains the identity element e and powers of $r^3$, to $r^{27}$. Thus the generator of H is then $r^3$. Would this be okay?
b) Since k divides n, we can write n as n = kp  for some p. Then we see that gcd(n,k) = k and thus:
|< r >| =  |r| =  n.
Then by Fundamental theorem of cyclic groups I can say that the group  has exactly one subgroup of order k, ie:  =  since n =  kp. And we are done.  
This is my first course in Group Theory so I'm rather shaky and insecure about my proofs. Your comments and help would be greatly appreciated.
 A: You are using $D_{30}$ to represent the dihedral group of order $60$, with $r$ corresponding to the "rotation", and some other element, call it $s$, the "reflection"; that is,
$$D_{30}=\left\langle r,s\Bigm| r^{30} = s^2 = 1,\quad sr = r^{-1}s\right\rangle.$$
Part (a): Your first part begins well enough, but it seems like you did not read the question carefully. The last part of part (a) asks you to find all generators of the subgroup $H$ you found.
You took $H=\langle r^3\rangle$. That's fine; this is a group of order $10$. But $r^3$ is not the only generator of this group. Perhaps you know that, in general, the cyclic group of order $n$, $C_n$ has $\varphi(n)$ generators, where $\varphi$ is Euler's function that counts the number of positive integers less than $n$ and relatively prime to $n$ (if you don't, then try to prove it). So $H$, being cyclic of order $10$, should have $\varphi(10) = 4$ generators. You've found one, there are three more to go.
Your part (b) also suffers a bit. You know that there is one and only one subgroup of $\langle r\rangle$ that has order $k$; that's fine. But you did not prove that every subgroup of $D_n$ of order $k$ must be a subgroup of $\langle r\rangle$! You need to show that too if you want your argument to hold. So you would need to show that the only elements of order $k$ all lie in $\langle r\rangle$; that's where you'll need the fact that $k\geq 3$. Perhaps you can show that every other element has other ideas about what their order is?
